Denote by the number of elements in satisfying . Frobenius’ theorem states that if is a divisor of , then is a multiple of .

In the article On an inverse Problem to Frobenius’ theorem (2011, Springerlink) Wei Meng and Jiangtao Shi propose the following problem.

Let be a positive integer. Classify all groups with the property that for every divisor of .

Denote by the largest positive integer such that for all divisors of . It is a standard result that if and only if is cyclic. Wei Meng and Jiangtao Shi have classified groups with and those with (with Kelin Chen).

Here are some of my thoughts about the general problem. Let be a group and suppose that for all . Let be a prime divisor of and let be a Sylow -subgroup of (of order ). Then

1) If , then is normal and cyclic.
2) If , then is normal or is cyclic.
3) If for , then is normal or
4) If , then is normal or else is cyclic and .

Denote by the number of Sylow -subgroups of . The facts above follow from a theorem of G. A. Miller which states that if , then and if , then .

Applying this we see that when for all , every Sylow subgroup of is cyclic or normal. Furthermore, any Sylow -subgroup is central if . To see this, note that is centralized by -sylows for since they are normal. The subgroup is also centralized by -sylows for since and the automorphism group of cyclic group of order has order coprime to . Hence by Burnside's normal complement theorem , where is cyclic and coprime to , and , reducing the classification of such to groups with order of the form .

A good starting point for classifying groups with for all might be to show that , where is cyclic and coprime to primes that are small enough. However, it seems difficult to find a good upper bound for the number of Sylow -subgroups in this case. Applying 1), 2), 3), 4) or Miller's theorem no longer works here, so improving Miller's result is one possibility. It might be true that . This is just a guess I made by looking at groups of order and values of when .

We would also need an upper bound for the size of primes dividing the order of the automorphism group of a -sylow . By 3), this would be a 2-group of order with a cyclic subgroup of order . I believe these are classified, so it would be a matter of computing the automorphism groups and thus giving an upper bound for the size of primes dividing . Then -sylows would centralize any -sylow for greater than any prime divisor of .

It would be easy to count the number of elements in -Sylow subgroups if they would always intersect trivially. However, this is not usually the case. For example, in you can find two different Sylow subgroups that have nontrivial intersection. The intersections don’t always have the same size either, as the group illustrates. Given a group with Sylow -subgroups of order , can we say anything about the intersections? The question is vague, but we can answer “nope, not without more information” as follows:

Let be a prime and integer. Then there exists a group with -Sylow subgroups of size such that: for any , we can find a pair of -Sylow subgroups of with .

We construct the group as a generalization of the example as follows. First, let be primes such that and . The existence of such is guaranteed by Dirichlet’s theorem. Unfortunately, this is a very non-trivial result from number theory. Some special cases such as and have elementary proofs though (you can adapt Euclid’s proof of the fact that there are infinitely many primes).

Now let be the nonabelian group of order , which exists since . Define the group as the direct product , where appears times in the product. A Sylow subgroup of then has order . Now there are different -Sylow subgroups in , so let be distinct Sylow subgroups of . Define

Then the are -Sylow subgroups of , and for all .

Hence in pairs of -Sylow subgroups can have all possible orders.

There is something we can say about the number of elements in Sylow -subgroups if we know the number of subgroups. First, if there is only only -Sylow subgroup, say of order , then obviously there are only elements. If there are several -Sylow subgroups, by Sylow’s theorem there are at least . It can be shown that if there are different -Sylow subgroups, then there are exactly elements in the subgroups. Furthermore, if there are more than different -Sylow subgroups, then there are more than elements in the subgroups. I think this result is due to G. A. Miller.

Suppose and are primes. Then according to Burnside’s theorem every group of order is solvable. We prove the special case here. The proof is very old, going back at least to the 1916 book Theory and Applications of Finite Groups by G. A. Miller, H. F. Blichfeldt and L. E. Dickson (Chapter VIII, §73, pg. 185).

Theorem: Every group of order is solvable.

We argue by induction. The case holds because groups of prime order are cyclic. Suppose then that the claim has been established for , and let be a group of order . It is enough to show that has a normal subgroup with and solvable, and hence by induction it suffices to find a nontrivial proper normal subgroup of . If has only one -Sylow subgroup, the subgroup is normal. Otherwise has exactly different -Sylow subgroups. Let be the intersection of two different -Sylow subgroups and such that has the largest possible order. Then since -groups satisfy the normalizer condition (ie. for proper subgroups, "normalizers grow") we get that

Next we show that cannot be a -group. If this were the case, then it would be contained in some -Sylow subgroup . Thus and by the maximality of this implies . Similarly so again by the maximality of we get , giving us the contradiction .

Thus by Cauchy’s theorem, there is an element of order in . Let be any -Sylow subgroup of . Since , the element does not normalize . Hence the subgroups are all distinct, and thus they are exactly all Sylow -subgroups of . Since normalizes , the subgroup is then contained in every Sylow -subgroup of . The intersection

of all -Sylow subgroups is always normal (the normal core of ). By induction, we may assume that it is trivial, so in particular we can assume that is trivial. Since we chose so that it has the largest possible order, the intersections of different -Sylow subgroups are trivial. Thus we get a total of nonidentity elements from the -Sylow subgroups. Therefore there can be only one -Sylow subgroup, which is thus a nontrivial proper normal subgroup.

Let be a prime. Then in any finite group of order a subgroup of index (in other words, a subgroup of order is normal. This is a pretty simple fact, and I will give outlines for four different proofs here. The first two are standard and the other two are not so common. Proof 3 I figured out myself and it is the most elementary one, requiring almost nothing besides the formula for the order of a product of two subgroups. Proof 4 I saw in an old paper (1895) by Frobenius (*).

Proof 1: In finite p-groups (more generally in nilpotent groups) it is true that for any proper subgroup , the normalizer is strictly larger than . Hence for a subgroup of order the normalizer must equal all of , implying that is normal.

Proof 2: For any finite group it is true that if a subgroup has index equal the smallest prime divisor of , the subgroup is normal. This can be seen by considering the permutation representation given by the coset action. Since is the smallest prime divisor of , the claim follows.

Proof 3: Let be a subgroup of index . Suppose that is not normal. Then there exists such that . Thus is equal to the product , but this is in contradiction with the following fact.

If is a subgroup of and such that , then .

Proof of fact: Since , we get and thus .

Proof 4: We proceed by induction. The case is clear. Since p-groups have nontrivial center, there exists of order . If , then by induction and thus . If , then and since is central.

The following proof that is simple is a slight improvement on the proof presented by Gallian in his Monthly article Another proof that is simple (1984). Before the proof we will need a few elementary lemmas.

Lemma 1. has elements of order , elements of order 3, and elements of order .

The elements of order are the -cycles, elements of order are -cycles, and elements of order are product of two transpositions.

Lemma 2. Let be a normal subgroup of a finite group . If and , then

Proof: The order of divides both and , so . Thus which implies .

An immediate corollary of this result is that a normal subgroup contains every element with order coprime to its index.

Lemma 3.The group has trivial center when .

Proof: Let , . Let be different elements from . Now , so . Therefore is in , as otherwise . Since also commutes with and , we see that is in and by the same argument. Thus , because the intersection of these three sets is . Therefore for any , and .

Lemma 4. A normal subgroup of order is central.

Proof: Suppose is a normal subgroup of order in . Then for any , we have . We cannot have because this would imply . Therefore , and .

Theorem: is simple.

Suppose that is a normal subgroup of . Now has order , so has order or by Lagrange’s theorem. If has order , then would contain elements of order since is coprime to . If has order or , then would contain all elements of order . If has order or , then would contain all elements of order . If has order , then would contain elements of order and elements of order . If has order , then is central since any normal subgroup of order is, but this isn’t actually possible since has trivial center. Therefore the only possibilities are that has order or order , ie. or .

Using Lagrange’s theorem, it is not difficult to show that every group of prime order is cyclic. If is a group of prime order, then each subgroup has order or . Thus is generated by any non-identity element.

Prime numbers are not the only numbers with this property. For example, any group of order is cyclic. To generalize this fact a bit, for and prime, and , a group of order is always cyclic. What about in general? Is there a simple arithmetic characterization for all cyclic numbers? This turns out to be true, and we have the following theorem, which we will prove next:

Theorem. Every group of order is cyclic if and only if .

Here denotes Euler’s totient function. The solution given here is a combination of the proof by Szele (1947) and another one given by Gallian and Moulton (1993). We will first need a lemma.

Lemma. If a finite group is not cyclic and every proper subgroup of is abelian, then has a nontrivial proper normal subgroup.

Proof. See the paper by Gallian and Moulton.

Since the lemma also holds for finite cyclic groups of non-prime order, we have the following corollary:

Corollary. If is a finite group of non-prime order with every proper subgroup abelian, then has a nontrivial proper normal subgroup.

Proof of theorem. We will first show by induction that forces any group of order to be cyclic. For the case the statement is clearly true. Suppose then that the claim is true for any integer less than and that . Let be a group of order . If is prime we are done, so we can assume that is not prime. Now has a prime factorization

and thus

This shows that , for being greater than one would imply that is a common divisor of and . Thus and are of the form

Therefore for any divisor of , we have . This shows that any proper subgroup of is cyclic, and thus has a nontrivial proper normal subgroup by the lemma. Since the order of also divides , the quotient group is also cyclic. Since and are abelian, they are solvable and thus is also solvable.

Let be a minimal normal subgroup of . Since is solvable, the normal subgroup has order for some prime . Then has order , and thus both and are cyclic. Now and for some elements and from . Notice that the order of must divide the order of , so has order or . If has order we are done, so suppose that has order .

Since is a normal subgroup, for some . Since has order , we notice that and thus . On the other hand, by Fermat’s little theorem. Since , we have . Thus and .

There is also another way to show that and commute. When is normal, there is a homomorphism with by mapping each element of to the corresponding conjugation map. Then divides , since a cyclic group of order has a exactly automorphisms. Thus since also divides . Therefore , which shows that is central.

Since and commute and have coprime order, the element has order . Therefore .

Proving the statement in the other direction is a bit easier. Suppose that . If is not squarefree, then has a prime divisor such that divides . Then is a noncyclic group of order . If is squarefree, then there exist prime divisors and of such that divides . But then a nonabelian group of order exists, and is a noncyclic group of order .

For a different proof, see for example the outline given by Dieter Jungnickel (1992).

After thinking about cyclic numbers, it is natural to generalize and wonder about other orders which force some property on a group. For a nice exposition of cyclic, abelian and nilpotent numbers, see “Nilpotent numbers” by Pakianathan and Shankar (2000).

These are all the examples I have found so far for numbers with no zero digit. We get more examples by taking any number with this property and multiplying it by a power of ten. For example, 36 = 3 * 6 + 3 * 6 and thus 360 = 3 * 60 + 3 * 60. This shows that there is an infinity of numbers with this property.

These facts are not too difficult to prove. We will prove the third equality here which is the most interesting one, as it has no zero in its decimal expansion. Numbers with zero are a bit more common, although there are not too many of either.

First, let’s express in form that is easier to manipulate. This way we easily find similar representations for , and .

Let be the amount of ‘s in this number. Now if , then , implying . What is as a fraction? Since , we get and so .

Thus . Then by multiplying by (which adds two zeros at the end) and adding (changing those zeros to ), we get . Therefore

And with the same method we find that

Summing these two together gives our result after a few simple manipulations:

as desired.

Are 36 = 3 * 6 + 3 * 6 and 1352 = 13 * 52 + 13 * 52 the only numbers that can be split this way from the middle?

Let be a proper subgroup of with index . We will show that . Left coset action gives us the homomorphism , where . Since is a proper subgroup and is simple, . By the first isomorphism theorem, . Therefore divides , implying that .

Next we will show that has a subgroup of index .

The amount of Sylow -subgroups must divide . Thus there are , , or Sylow -subgroups. If there is only one Sylow -subgroup, it is normal. Thus there has to be more than one Sylow -subgroup, since is simple. The amount of Sylow -subgroups is equal to the index of the normalizer of a Sylow -subgroup, so is not a possibility as shown before. If there are Sylow 2-subgroups, then the normalizer of a Sylow -subgroup is of index .

Suppose then that there are Sylow 2-subgroups in . It is not difficult to show that there are Sylow 5-subgroups in total, giving us different elements. Thus there exist two different Sylow 2-subgroups and with nontrivial intersection. Otherwise we would find different elements, which is not possible. Since and are Abelian, the intersection is normal in them both. Thus it is also normal in the subgroup generated by and . Since is simple, this subgroup is proper. As shown in the first paragraph, , and thus . By Lagrange’s theorem, divides the order of . Thus , because we also have . Then is a subgroup of index .

We have now shown that there exists some with . The left coset action gives us a homomorphism with . By the same argument as before, . Denote . Then . Now is normal in , because it has order and therefore index . Then is normal in . We know that is simple. Furthermore, any subgroup of with more than two elements always contains an even element other than the identity. These two facts imply that , and so . Since and both have the same order, .

Lemma 3. If is a group and is an automorphism of , then for every of finite order we have .

Proof: Follows from the fact that if and only if .

Lemma 4. Any automorphism preserves the parity of permutations.

Proof: Since the commutator subgroup of is , we have that is a characteristic subgroup of . Thus , and so .

for , and .

Proof: In general, this result is true for every with and .

Because is generated by the elements and , we have that any automorphism is completely determined by and . By lemma 3, there are choices for and 2 choices for , so there can be at most different automorphisms. Since , we have that and so .

Similarly is generated by and . By lemma 3 and lemma 4, we have choices for and choices for . Thus there are at most automorphisms of . On the other hand, , so is a multiple of by Lagrange’s theorem. Thus there are automorphisms, and . Therefore .

The case of is similar. Now is generated by and and there are choices for and choices for . As before, , so is a multiple of . Thus there are or automorphisms. We will show that one of the choices for and is not possible, which then shows that there are automorphisms, and thus and as before. Suppose is an automorphism with and . Then

But this is not possible, as has order and order . Thus there are less than automorphisms, and so .

This approach does not work with larger symmetric groups. The next symmetric group for which the claim is true is . In , we would have 126 choices for and 720 choices for . This gives us a total of possibilities, so a different method is required.

Let be a group. For a nonempty subset of and element , define the coset by:

We know from Lagrange’s theorem that if is a subgroup, then the cosets partition into disjoint subsets. The question is: are there any other subsets which partition with cosets? The following result answers the question completely, and we will prove it here.

The cosets of partition if and only if is a coset of a subgroup of .

Suppose first that the cosets of partition . Let be the coset that contains the neutral element . Then if , we also have . Thus , as we assumed that the cosets induce a disjoint partition of . Now , so for some . This means that . Finally, for all , we get . This shows that , a coset of , is a subgroup of . Since , we get , and thus is a coset of the subgroup .

For the converse, suppose , where is a subgroup of . Then , so the cosets of and are the same. Since is a subgroup, the cosets partition by Lagrange’s theorem.

The following exercise can be found in I.N. Herstein’s book Topics in Algebra (second edition, pg. 53):

Give an example of a group , subgroup and an element such that but .

First we notice that cannot be normal in . This means that cannot be Abelian. Secondly, must be infinite. Otherwise would imply , since and always have the same order. This shows that we should be looking for an infinite, nonabelian group with an infinite subgroup that has the desired property. Here is my example:

Let , the set of bijections from to . Let . Here does not include zero. Now is a subgroup of . Define the map by for all .

We will prove that for all , the commutator subgroup of (denoted ) is equal to , the the alternating group of degree .

First we will show that any 3-cycle must be in the commutator subgroup. Let be a 3-cycle from . Now

Therefore is a commutator, and thus is in the commutator subgroup. Since is generated by 3-cycles when , the commutator subgroup contains all of .

Next, note that any commutator , with and permutations from , must be an even permutation. It is known that has the same parity as and has the same parity as , and thus has the same parity as . This shows that their product must be even.

Thus every permutation in the commutator subgroup is even. Since the commutator subgroup contains , the group of even permutations, it must be equal to .

When , the alternating group is simple. From this fact and what we just proved it follows that is not solvable when .

Let be a group, and a nonempty subset of . The following statements are equivalent:

is a subgroup of

For all : and

For all :

For all :

if and only if

If , then

If and , then

We will equivalence of being a subgroup and statement for granted and use it to prove statement and equivalent to being a subgroup.

Proof of statement 6: Suppose first that is a subgroup of . Let . Now for all , so . Let . Then , so . Thus .

For the other direction, suppose that always implies . For any we get so for . Then , so . Similarly, so for some . Now , so . Finally, for all , we have . Thus is a subgroup of .

Proof of statement 7: If is a subgroup, then when and , we cannot have . If this would happen, then since .

Suppose then that and implies in the nonempty subset . First we will show that . Let be some element of . If we had , then , which is a contradiction. Therefore . If we had , then , again a contradiction. Thus is closed under inverses. Suppose and . Then . Thus if we had , we would have , a contradiction. Therefore .